Ex 2.2 sets ncert maths solution class 11

      EXERCISE 2.2 (Class 11)

1. Let A = {1, 2, 3, … , 14}. Define a relation R from A to A by R = {(xy): 3x – y = 0, where xy ∈ A}. Write down its domain, codomain and range.

Solution: The relation R from A to A is given as:

R = {(xy): 3x – y = 0, where xy ∈ A}

= {(xy): 3x = y, where xy ∈ A}

Hence,

R = {(1, 3), (2, 6), (3, 9), (4, 12)}

Now,

Hence, Domain of R = {1, 2, 3, 4}

The whole set A is the codomain of the R.

Hence, Codomain of R  = {1, 2, 3, …, 14} = A

The range of R is the set of all second elements of the ordered pairs in the relation.

Hence, Range of R = {3, 6, 9, 12}

2. Define a relation R on the set N of natural numbers by R = {(xy): y = x + 5, x is a natural number less than 4; xy ∈ N}. Depict this relationship using roster form. Write down the domain and the range.

Solution: The relation R is given by:

R = {(xy): y = x + 5, x is a natural number less than 4, xy ∈ N}

The natural numbers less than 4 ={1, 2, 3}

R = {(1, 6), (2, 7), (3, 8)}

Now,

The domain of R is the set of all first elements of the ordered pairs in R.

Hence, Domain of R = {1, 2, 3}

The range of R is the set of all second elements of the ordered pairs in R

Hence, Range of R = {6, 7, 8}

3. A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(xy): the difference between x and y is odd; x ∈ A, ∈ B}. Write R in roster form.

Solution: Given,

A = {1, 2, 3, 5} and B = {4, 6, 9}

The relation from A to B is given as

R = {(xy): the difference between x and y is odd; x ∈ A, ∈ B}

R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

4. The figure shows a relationship between the sets P and Q. Write this relation

(i) in set-builder form (ii) in roster form

What is its domain and range?

NCERT Solutions Class 11 Mathematics Chapter 2 ex.2.2 - 1

Solution: From the figure, it’s seen that

P = {5, 6, 7}, Q = {3, 4, 5}

The relation between P and Q:

Set-builder form

(i) R = {(x, y): y = x – 2; x ∈ P}

Roster form

(ii) R = {(5, 3), (6, 4), (7, 5)}

Domain of R = {5, 6, 7}

Range of R = {3, 4, 5}

5. Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by

{(ab): ab ∈ A, b is exactly divisible by a}.

(i) Write R in roster form

(ii) Find the domain of R

(iii) Find the range of R

Solution: Given,

A = {1, 2, 3, 4, 6} and

relation R = {(ab): ab ∈ A, b is exactly divisible by a}

Hence,

(i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}

(ii) Domain of R = {1, 2, 3, 4, 6}

(iii) Range of R = {1, 2, 3, 4, 6}

6. Determine the domain and range of the relation R defined by R = {(xx + 5): x ∈ {0, 1, 2, 3, 4, 5}}.

Solution: Given,

Relation R = {(xx + 5): x ∈ {0, 1, 2, 3, 4, 5}}

R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}

So,

Domain of R = {0, 1, 2, 3, 4, 5} and,

Range of R = {5, 6, 7, 8, 9, 10}

7. Write the relation R = {(xx3): is a prime number less than 10} in roster form.

Solution: Given,

Relation R = {(xx3): is a prime number less than 10}

The prime numbers less than 10 = {2, 3, 5, 7}.

Therefore,  R = {(2, 8), (3, 27), (5, 125), (7, 343)}

8. Let A = {xy, z} and B = {1, 2}. Find the number of relations from A to B.

Solution:  Given, A = {xy, z} and B = {1, 2}

Now,

A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}

As n(A × B) = 6,

the number of subsets of A × B  =  26.

Thus, the number of relations from A to B =  26.

9. Let R be the relation on Z defined by R = {(ab): ab ∈ Z, – b is an integer}. Find the domain and range of R.

Solution: Given,

Relation R = {(ab): ab ∈ Z, – b is an integer}

We know that the difference between any two integers is always an integer.

Therefore,

Domain of R = Z

Range of R = Z.


Ex 2.1 sets ncert maths solution class 11

Class 12 relation and functions multiple choice

Case study relation and function 3
An organization conducted bike race under 2 different categories- boys and girls.

 

 

Team Gmath

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